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Electric dipole and Gauss' Law

  1. Mar 17, 2007 #1
    We have an electric dipole. Now, let us draw a Gaussian surface around our electric dipole. Now, the total charge enclosed by our Gaussian surface is zero, so according the Gauss' Law the flux through the Gaussian surface is zero, and so is the electric field intensity due the electric dipole.

    But, when we apply Coulomb's Law, we get an expression for electric field intensity at a point due to an electric dipole.
    So, my question is-

    Am I going wrong somewhere in applying the Gauss' theorem?
    If not, why are we getting this difference in the solution to this problem?
     
  2. jcsd
  3. Mar 17, 2007 #2
    net flux=0 does not imply field strength=0 at a particular point.
     
  4. Mar 18, 2007 #3

    jtbell

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    Staff: Mentor

    No... the flux through the Gaussian surface is zero, and so the net flux outwards (through some portions of the surface) equals the net flux inwards (through other portions of the surface).
     
  5. Mar 18, 2007 #4
    [tex]\int_{0}^{2\pi}{\sin\left(x\right)}dx = 0,[/tex] but that does not imply sin(x) is zero for all values in the interval.
     
  6. Mar 18, 2007 #5
    So, then how do we find the electric field intensity at a point due to an electric dipole using the Gauss' Law?
     
  7. Mar 18, 2007 #6
    Its not easy to calculate E.Field due to dipole using Gauss Law. It is because you'll have to choose a gaussian surface such a way that you are able to calculate the E.Field there. Remember Gauss's law basically tells about the flux and *not* of E.Field.
     
  8. Mar 18, 2007 #7

    jtbell

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    Staff: Mentor

    I don't think you can, at least not directly. At least I've never seen it done. In order to use Gauss's Law to find the electric field, the arrangement of charge has to be symmetric enough that you can infer the direction of the electric field at all points. Then you construct a Gaussian surface which makes evaluating the integral easy. A dipole doesn't have enough symmetry for this.

    However, you can of course find the field of each charge individually, using Gauss's Law, which of course gives you Coulomb's Law. Then add the two fields using the principle of superposition.

    All the situations where I've seen Gauss's Law used to find the electric field have either planar, cylindrical or spherical symmetry.
     
  9. Mar 18, 2007 #8
    You can use Gauss' Law only in places where you have symmetric uniform charge distribution as in case of dipole there is no symmetry so you cannot .Remember we use Gauss' law for Sphere's Cylinder's because there is Uniform charge distribution and Symmetry.
     
  10. Mar 19, 2007 #9
    Thank you for helping.
     
  11. Apr 13, 2007 #10
    According to de broglie relation lambda=h/mv ...which implies that vrlocity is inversely proportional to wavelength. But According to the reletion

    V=n lambda ... velocity is directly proportional to wavelength...... How That diffenence is Causesd ? Am i going wrong Somowhere ?
     
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